Limit of $argmax$ equals $argmax$ of limit?
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Let $X$ be some set such as $a,b,c$ or $mathbb R^n$. We want to choose a vector $x=(x_0,x_1,...)in X^infty$ that maximizes the sum below. Interpret this as a value $x_t$ for each time period $t$. (Assume the sum exists for all $x$).
My question is, under what conditions can we "move the argmax inside the limit operator", as follows?
$$argmax_xin X^inftylim_Tto inftysum_t=0^T gamma^tf(x_t)=lim_Tto inftyargmax_xin X^inftysum_t=0^T gamma^tf(x_t)$$
With $gammain (0,1)$. This is essentially a "limit of a set" equation. I'm not sure where to start to find out an answer to this question.
EDIT: I don't necessarily want to assume that $f:Xto mathbb R$ is continuous, or even that $X$ is an infinite set.
EDIT: I just realized that I mistyped the equation at first... I was too hasty, and now understand people's comments... I forgot to add the $gamma$. Sorry for wasting people's time.
limits
asked Sep 7 at 3:54
Programmer2134
3,25821048
 |Â
show 7 more comments
up vote
2
down vote
favorite
Let $X$ be some set such as $a,b,c$ or $mathbb R^n$. We want to choose a vector $x=(x_0,x_1,...)in X^infty$ that maximizes the sum below. Interpret this as a value $x_t$ for each time period $t$. (Assume the sum exists for all $x$).
My question is, under what conditions can we "move the argmax inside the limit operator", as follows?
$$argmax_xin X^inftylim_Tto inftysum_t=0^T gamma^tf(x_t)=lim_Tto inftyargmax_xin X^inftysum_t=0^T gamma^tf(x_t)$$
With $gammain (0,1)$. This is essentially a "limit of a set" equation. I'm not sure where to start to find out an answer to this question.
EDIT: I don't necessarily want to assume that $f:Xto mathbb R$ is continuous, or even that $X$ is an infinite set.
EDIT: I just realized that I mistyped the equation at first... I was too hasty, and now understand people's comments... I forgot to add the $gamma$. Sorry for wasting people's time.
limits
asked Sep 7 at 3:54
Programmer2134
3,25821048
What is exactly $operatornamearg max$?
– Guido A.
Sep 7 at 3:56
@GuidoA. $operatornamearg,maxf(x)$ is a point $x_0$ at which $f$ is maximal. Saludos de Alemana a UBA. ;-)
– amsmath
Sep 7 at 4:09
1
@amsmath, actually $argmax f(x)$ is the set of all such points.
– Programmer2134
Sep 7 at 4:19
@Programmer2134 That is indeed the most accurate definition.
– amsmath
Sep 7 at 4:21
1
@amsmath Grüße :P
– Guido A.
Sep 7 at 4:27
 |Â
show 7 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $X$ be some set such as $a,b,c$ or $mathbb R^n$. We want to choose a vector $x=(x_0,x_1,...)in X^infty$ that maximizes the sum below. Interpret this as a value $x_t$ for each time period $t$. (Assume the sum exists for all $x$).
My question is, under what conditions can we "move the argmax inside the limit operator", as follows?
$$argmax_xin X^inftylim_Tto inftysum_t=0^T gamma^tf(x_t)=lim_Tto inftyargmax_xin X^inftysum_t=0^T gamma^tf(x_t)$$
With $gammain (0,1)$. This is essentially a "limit of a set" equation. I'm not sure where to start to find out an answer to this question.
EDIT: I don't necessarily want to assume that $f:Xto mathbb R$ is continuous, or even that $X$ is an infinite set.
EDIT: I just realized that I mistyped the equation at first... I was too hasty, and now understand people's comments... I forgot to add the $gamma$. Sorry for wasting people's time.
limits
asked Sep 7 at 3:54
Programmer2134
3,25821048
Let $X$ be some set such as $a,b,c$ or $mathbb R^n$. We want to choose a vector $x=(x_0,x_1,...)in X^infty$ that maximizes the sum below. Interpret this as a value $x_t$ for each time period $t$. (Assume the sum exists for all $x$).
My question is, under what conditions can we "move the argmax inside the limit operator", as follows?
$$argmax_xin X^inftylim_Tto inftysum_t=0^T gamma^tf(x_t)=lim_Tto inftyargmax_xin X^inftysum_t=0^T gamma^tf(x_t)$$
With $gammain (0,1)$. This is essentially a "limit of a set" equation. I'm not sure where to start to find out an answer to this question.
EDIT: I don't necessarily want to assume that $f:Xto mathbb R$ is continuous, or even that $X$ is an infinite set.
EDIT: I just realized that I mistyped the equation at first... I was too hasty, and now understand people's comments... I forgot to add the $gamma$. Sorry for wasting people's time.
limits
limits
asked Sep 7 at 3:54
Programmer2134
3,25821048
asked Sep 7 at 3:54
Programmer2134
3,25821048
edited Sep 8 at 7:27
asked Sep 7 at 3:54
Programmer2134
3,25821048
asked Sep 7 at 3:54
Programmer2134
3,25821048
asked Sep 7 at 3:54
Programmer2134
3,25821048
3,25821048
What is exactly $operatornamearg max$?
– Guido A.
Sep 7 at 3:56
@GuidoA. $operatornamearg,maxf(x)$ is a point $x_0$ at which $f$ is maximal. Saludos de Alemana a UBA. ;-)
– amsmath
Sep 7 at 4:09
1
@amsmath, actually $argmax f(x)$ is the set of all such points.
– Programmer2134
Sep 7 at 4:19
@Programmer2134 That is indeed the most accurate definition.
– amsmath
Sep 7 at 4:21
1
@amsmath Grüße :P
– Guido A.
Sep 7 at 4:27
 |Â
show 7 more comments
What is exactly $operatornamearg max$?
– Guido A.
Sep 7 at 3:56
@GuidoA. $operatornamearg,maxf(x)$ is a point $x_0$ at which $f$ is maximal. Saludos de Alemana a UBA. ;-)
– amsmath
Sep 7 at 4:09
1
@amsmath, actually $argmax f(x)$ is the set of all such points.
– Programmer2134
Sep 7 at 4:19
@Programmer2134 That is indeed the most accurate definition.
– amsmath
Sep 7 at 4:21
1
@amsmath Grüße :P
– Guido A.
Sep 7 at 4:27
What is exactly $operatornamearg max$?
– Guido A.
Sep 7 at 3:56
What is exactly $operatornamearg max$?
– Guido A.
Sep 7 at 3:56
@GuidoA. $operatornamearg,maxf(x)$ is a point $x_0$ at which $f$ is maximal. Saludos de Alemana a UBA. ;-)
– amsmath
Sep 7 at 4:09
@GuidoA. $operatornamearg,maxf(x)$ is a point $x_0$ at which $f$ is maximal. Saludos de Alemana a UBA. ;-)
– amsmath
Sep 7 at 4:09
1
1
@amsmath, actually $argmax f(x)$ is the set of all such points.
– Programmer2134
Sep 7 at 4:19
@amsmath, actually $argmax f(x)$ is the set of all such points.
– Programmer2134
Sep 7 at 4:19
@Programmer2134 That is indeed the most accurate definition.
– amsmath
Sep 7 at 4:21
@Programmer2134 That is indeed the most accurate definition.
– amsmath
Sep 7 at 4:21
1
1
@amsmath Grüße :P
– Guido A.
Sep 7 at 4:27
@amsmath Grüße :P
– Guido A.
Sep 7 at 4:27
 |Â
show 7 more comments
1 Answer
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0
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For finite sums, if $max_xin X^T sum_t=1^T-1 gamma^t f(x_t)$ exists, we must have that $f$ assumes its maximum $M$ in a set $N=argmax_xin X f$. The expression $sum_t=0^T-1 gamma^t f(x_t)$ is maximal if each of the terms is maximal, so $argmax_xin X^Tsum_t=1^T-1 gamma^t f(x_t)=N^T$ (if any of the entries is not in the argmax, then you can find a way to make the sum larger).
Now we consider the infinite sum. As $f(x)le M$ on $X$, we have
$$sum_t=0^infty gamma^t f(x_t) le Msum_t=0^infty gamma^t=fracM1-gamma.
$$
If we have a sequence $(x_t)$ such that $f(x_t)=M$ for every $t$, we have equality. If $f(x_t)<M$ for one $t$, then $sum_t=0^infty gamma^t f(x_t) < M$. So
$$argmax_xin X^infty sum_t=0^infty gamma^t f(x_t) = N^infty.
$$
At the same time, as the finite sum only cares about the first terms,
$$argmax_xin X^infty sum_t=0^T-1 gamma^t f(x_t) = N^Ttimes prod_t=T^infty X simeq N^Ttimes X^infty.
$$
All we need now is a sense of convergence of sets where $N^Ttimes X^inftyto N^infty$ as $Ttoinfty$. My set / category theory are rusty, but I think this is an inverse limit that we can write as the intersection
$$N^infty = bigcap_T=0^infty (x_0,dots, x_T,y_T+1, y_T+2, dots) : x_i in N, y_jin Xsubset X^infty.
$$
answered Sep 7 at 18:13
Kusma
3,415218
I can't really follow this, but you are assuming that $X=mathbb R$ it seems to me. But $X$ could really be anything. e.g. it could be $text John Travolta, textCheese, 5$
– Programmer2134
Sep 8 at 5:46
No, I am assuming that $X$ is any set and $f:XtomathbbR$. The supremum I take is over a set of values of $f$, which is a set of real numbers. I show that for both sides of your limit to be defined, $f$ has to be nonpositive with a nonempty zero set, and we can explicitly calculate the argmax in that case.
– Kusma
Sep 8 at 5:52
sorry, I now see that I've made a very clumsy mistake in the question... Very sorry for this.
– Programmer2134
Sep 8 at 7:28
@Programmer2134 no worries, the answer is almost the same and I think I improved my exposition :)
– Kusma
Sep 8 at 8:52
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
For finite sums, if $max_xin X^T sum_t=1^T-1 gamma^t f(x_t)$ exists, we must have that $f$ assumes its maximum $M$ in a set $N=argmax_xin X f$. The expression $sum_t=0^T-1 gamma^t f(x_t)$ is maximal if each of the terms is maximal, so $argmax_xin X^Tsum_t=1^T-1 gamma^t f(x_t)=N^T$ (if any of the entries is not in the argmax, then you can find a way to make the sum larger).
Now we consider the infinite sum. As $f(x)le M$ on $X$, we have
$$sum_t=0^infty gamma^t f(x_t) le Msum_t=0^infty gamma^t=fracM1-gamma.
$$
If we have a sequence $(x_t)$ such that $f(x_t)=M$ for every $t$, we have equality. If $f(x_t)<M$ for one $t$, then $sum_t=0^infty gamma^t f(x_t) < M$. So
$$argmax_xin X^infty sum_t=0^infty gamma^t f(x_t) = N^infty.
$$
At the same time, as the finite sum only cares about the first terms,
$$argmax_xin X^infty sum_t=0^T-1 gamma^t f(x_t) = N^Ttimes prod_t=T^infty X simeq N^Ttimes X^infty.
$$
All we need now is a sense of convergence of sets where $N^Ttimes X^inftyto N^infty$ as $Ttoinfty$. My set / category theory are rusty, but I think this is an inverse limit that we can write as the intersection
$$N^infty = bigcap_T=0^infty (x_0,dots, x_T,y_T+1, y_T+2, dots) : x_i in N, y_jin Xsubset X^infty.
$$
answered Sep 7 at 18:13
Kusma
3,415218
I can't really follow this, but you are assuming that $X=mathbb R$ it seems to me. But $X$ could really be anything. e.g. it could be $text John Travolta, textCheese, 5$
– Programmer2134
Sep 8 at 5:46
No, I am assuming that $X$ is any set and $f:XtomathbbR$. The supremum I take is over a set of values of $f$, which is a set of real numbers. I show that for both sides of your limit to be defined, $f$ has to be nonpositive with a nonempty zero set, and we can explicitly calculate the argmax in that case.
– Kusma
Sep 8 at 5:52
sorry, I now see that I've made a very clumsy mistake in the question... Very sorry for this.
– Programmer2134
Sep 8 at 7:28
@Programmer2134 no worries, the answer is almost the same and I think I improved my exposition :)
– Kusma
Sep 8 at 8:52
add a comment |Â
up vote
0
down vote
For finite sums, if $max_xin X^T sum_t=1^T-1 gamma^t f(x_t)$ exists, we must have that $f$ assumes its maximum $M$ in a set $N=argmax_xin X f$. The expression $sum_t=0^T-1 gamma^t f(x_t)$ is maximal if each of the terms is maximal, so $argmax_xin X^Tsum_t=1^T-1 gamma^t f(x_t)=N^T$ (if any of the entries is not in the argmax, then you can find a way to make the sum larger).
Now we consider the infinite sum. As $f(x)le M$ on $X$, we have
$$sum_t=0^infty gamma^t f(x_t) le Msum_t=0^infty gamma^t=fracM1-gamma.
$$
If we have a sequence $(x_t)$ such that $f(x_t)=M$ for every $t$, we have equality. If $f(x_t)<M$ for one $t$, then $sum_t=0^infty gamma^t f(x_t) < M$. So
$$argmax_xin X^infty sum_t=0^infty gamma^t f(x_t) = N^infty.
$$
At the same time, as the finite sum only cares about the first terms,
$$argmax_xin X^infty sum_t=0^T-1 gamma^t f(x_t) = N^Ttimes prod_t=T^infty X simeq N^Ttimes X^infty.
$$
All we need now is a sense of convergence of sets where $N^Ttimes X^inftyto N^infty$ as $Ttoinfty$. My set / category theory are rusty, but I think this is an inverse limit that we can write as the intersection
$$N^infty = bigcap_T=0^infty (x_0,dots, x_T,y_T+1, y_T+2, dots) : x_i in N, y_jin Xsubset X^infty.
$$
answered Sep 7 at 18:13
Kusma
3,415218
I can't really follow this, but you are assuming that $X=mathbb R$ it seems to me. But $X$ could really be anything. e.g. it could be $text John Travolta, textCheese, 5$
– Programmer2134
Sep 8 at 5:46
No, I am assuming that $X$ is any set and $f:XtomathbbR$. The supremum I take is over a set of values of $f$, which is a set of real numbers. I show that for both sides of your limit to be defined, $f$ has to be nonpositive with a nonempty zero set, and we can explicitly calculate the argmax in that case.
– Kusma
Sep 8 at 5:52
sorry, I now see that I've made a very clumsy mistake in the question... Very sorry for this.
– Programmer2134
Sep 8 at 7:28
@Programmer2134 no worries, the answer is almost the same and I think I improved my exposition :)
– Kusma
Sep 8 at 8:52
add a comment |Â
up vote
0
down vote
up vote
0
down vote
For finite sums, if $max_xin X^T sum_t=1^T-1 gamma^t f(x_t)$ exists, we must have that $f$ assumes its maximum $M$ in a set $N=argmax_xin X f$. The expression $sum_t=0^T-1 gamma^t f(x_t)$ is maximal if each of the terms is maximal, so $argmax_xin X^Tsum_t=1^T-1 gamma^t f(x_t)=N^T$ (if any of the entries is not in the argmax, then you can find a way to make the sum larger).
Now we consider the infinite sum. As $f(x)le M$ on $X$, we have
$$sum_t=0^infty gamma^t f(x_t) le Msum_t=0^infty gamma^t=fracM1-gamma.
$$
If we have a sequence $(x_t)$ such that $f(x_t)=M$ for every $t$, we have equality. If $f(x_t)<M$ for one $t$, then $sum_t=0^infty gamma^t f(x_t) < M$. So
$$argmax_xin X^infty sum_t=0^infty gamma^t f(x_t) = N^infty.
$$
At the same time, as the finite sum only cares about the first terms,
$$argmax_xin X^infty sum_t=0^T-1 gamma^t f(x_t) = N^Ttimes prod_t=T^infty X simeq N^Ttimes X^infty.
$$
All we need now is a sense of convergence of sets where $N^Ttimes X^inftyto N^infty$ as $Ttoinfty$. My set / category theory are rusty, but I think this is an inverse limit that we can write as the intersection
$$N^infty = bigcap_T=0^infty (x_0,dots, x_T,y_T+1, y_T+2, dots) : x_i in N, y_jin Xsubset X^infty.
$$
answered Sep 7 at 18:13
Kusma
3,415218
For finite sums, if $max_xin X^T sum_t=1^T-1 gamma^t f(x_t)$ exists, we must have that $f$ assumes its maximum $M$ in a set $N=argmax_xin X f$. The expression $sum_t=0^T-1 gamma^t f(x_t)$ is maximal if each of the terms is maximal, so $argmax_xin X^Tsum_t=1^T-1 gamma^t f(x_t)=N^T$ (if any of the entries is not in the argmax, then you can find a way to make the sum larger).
Now we consider the infinite sum. As $f(x)le M$ on $X$, we have
$$sum_t=0^infty gamma^t f(x_t) le Msum_t=0^infty gamma^t=fracM1-gamma.
$$
If we have a sequence $(x_t)$ such that $f(x_t)=M$ for every $t$, we have equality. If $f(x_t)<M$ for one $t$, then $sum_t=0^infty gamma^t f(x_t) < M$. So
$$argmax_xin X^infty sum_t=0^infty gamma^t f(x_t) = N^infty.
$$
At the same time, as the finite sum only cares about the first terms,
$$argmax_xin X^infty sum_t=0^T-1 gamma^t f(x_t) = N^Ttimes prod_t=T^infty X simeq N^Ttimes X^infty.
$$
All we need now is a sense of convergence of sets where $N^Ttimes X^inftyto N^infty$ as $Ttoinfty$. My set / category theory are rusty, but I think this is an inverse limit that we can write as the intersection
$$N^infty = bigcap_T=0^infty (x_0,dots, x_T,y_T+1, y_T+2, dots) : x_i in N, y_jin Xsubset X^infty.
$$
answered Sep 7 at 18:13
Kusma
3,415218
edited Sep 8 at 8:51
answered Sep 7 at 18:13
Kusma
3,415218
answered Sep 7 at 18:13
Kusma
3,415218
answered Sep 7 at 18:13
Kusma
3,415218
3,415218
I can't really follow this, but you are assuming that $X=mathbb R$ it seems to me. But $X$ could really be anything. e.g. it could be $text John Travolta, textCheese, 5$
– Programmer2134
Sep 8 at 5:46
No, I am assuming that $X$ is any set and $f:XtomathbbR$. The supremum I take is over a set of values of $f$, which is a set of real numbers. I show that for both sides of your limit to be defined, $f$ has to be nonpositive with a nonempty zero set, and we can explicitly calculate the argmax in that case.
– Kusma
Sep 8 at 5:52
sorry, I now see that I've made a very clumsy mistake in the question... Very sorry for this.
– Programmer2134
Sep 8 at 7:28
@Programmer2134 no worries, the answer is almost the same and I think I improved my exposition :)
– Kusma
Sep 8 at 8:52
add a comment |Â
I can't really follow this, but you are assuming that $X=mathbb R$ it seems to me. But $X$ could really be anything. e.g. it could be $text John Travolta, textCheese, 5$
– Programmer2134
Sep 8 at 5:46
No, I am assuming that $X$ is any set and $f:XtomathbbR$. The supremum I take is over a set of values of $f$, which is a set of real numbers. I show that for both sides of your limit to be defined, $f$ has to be nonpositive with a nonempty zero set, and we can explicitly calculate the argmax in that case.
– Kusma
Sep 8 at 5:52
sorry, I now see that I've made a very clumsy mistake in the question... Very sorry for this.
– Programmer2134
Sep 8 at 7:28
@Programmer2134 no worries, the answer is almost the same and I think I improved my exposition :)
– Kusma
Sep 8 at 8:52
I can't really follow this, but you are assuming that $X=mathbb R$ it seems to me. But $X$ could really be anything. e.g. it could be $text John Travolta, textCheese, 5$
– Programmer2134
Sep 8 at 5:46
I can't really follow this, but you are assuming that $X=mathbb R$ it seems to me. But $X$ could really be anything. e.g. it could be $text John Travolta, textCheese, 5$
– Programmer2134
Sep 8 at 5:46
No, I am assuming that $X$ is any set and $f:XtomathbbR$. The supremum I take is over a set of values of $f$, which is a set of real numbers. I show that for both sides of your limit to be defined, $f$ has to be nonpositive with a nonempty zero set, and we can explicitly calculate the argmax in that case.
– Kusma
Sep 8 at 5:52
No, I am assuming that $X$ is any set and $f:XtomathbbR$. The supremum I take is over a set of values of $f$, which is a set of real numbers. I show that for both sides of your limit to be defined, $f$ has to be nonpositive with a nonempty zero set, and we can explicitly calculate the argmax in that case.
– Kusma
Sep 8 at 5:52
sorry, I now see that I've made a very clumsy mistake in the question... Very sorry for this.
– Programmer2134
Sep 8 at 7:28
sorry, I now see that I've made a very clumsy mistake in the question... Very sorry for this.
– Programmer2134
Sep 8 at 7:28
@Programmer2134 no worries, the answer is almost the same and I think I improved my exposition :)
– Kusma
Sep 8 at 8:52
@Programmer2134 no worries, the answer is almost the same and I think I improved my exposition :)
– Kusma
Sep 8 at 8:52
add a comment |Â
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What is exactly $operatornamearg max$?
– Guido A.
Sep 7 at 3:56
@GuidoA. $operatornamearg,maxf(x)$ is a point $x_0$ at which $f$ is maximal. Saludos de Alemana a UBA. ;-)
– amsmath
Sep 7 at 4:09
1
@amsmath, actually $argmax f(x)$ is the set of all such points.
– Programmer2134
Sep 7 at 4:19
@Programmer2134 That is indeed the most accurate definition.
– amsmath
Sep 7 at 4:21
1
@amsmath Grüße :P
– Guido A.
Sep 7 at 4:27